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If the conjecture is false, then there exists a sequence with no arithmetic progressions of length 3 such that the sum of the reciprocals of the series diverges.
Timefly wrote:I'm not sure there's any basis for the following statement...If the conjecture is false, then there exists a sequence with no arithmetic progressions of length 3 such that the sum of the reciprocals of the series diverges.
It could be that in every sum of the reciprocals of a series that diverges there is always an arithmetic progression of length 3 but not always one of length 4.
Correct me if I'm mistaken.
tomtom2357 wrote:No matter what sequence you have that has no arithmetic progression, it will converge, and it will converge to a number at or below a specific constant. I will call this (hypothetical) constant c.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l wrote:tomtom2357 wrote:No matter what sequence you have that has no arithmetic progression, it will converge, and it will converge to a number at or below a specific constant. I will call this (hypothetical) constant c.
Do you have any reason why that should be true? As an analogy, the sum of 1/r^n converges for any r less than 1, but there is no upper bound on what this can converge to. Finding such a c would indeed prove the conjecture, but the existence of such a c is a stronger statement than the conjecture itself.
tomtom2357 wrote:mike-l wrote:tomtom2357 wrote:No matter what sequence you have that has no arithmetic progression, it will converge, and it will converge to a number at or below a specific constant. I will call this (hypothetical) constant c.
Do you have any reason why that should be true? As an analogy, the sum of 1/r^n converges for any r less than 1, but there is no upper bound on what this can converge to. Finding such a c would indeed prove the conjecture, but the existence of such a c is a stronger statement than the conjecture itself.
You are right if I extend this to arithmetic progressions of length more than three, as you can take infinitely many sequences with the limit sequence being {1,2,3,...} and since this goes to infinity (as in the sum of the reciprocals diverges), so does the limit of the sequences. However, I am trying to solve the problem for arithmetic progressions of length three, so this problem doesn't arise.
I have found that (using a shanks transformation of the series) that the series converges to about 3.0079
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l wrote:tomtom2357 wrote:mike-l wrote:tomtom2357 wrote:No matter what sequence you have that has no arithmetic progression, it will converge, and it will converge to a number at or below a specific constant. I will call this (hypothetical) constant c.
Do you have any reason why that should be true? As an analogy, the sum of 1/r^n converges for any r less than 1, but there is no upper bound on what this can converge to. Finding such a c would indeed prove the conjecture, but the existence of such a c is a stronger statement than the conjecture itself.
You are right if I extend this to arithmetic progressions of length more than three, as you can take infinitely many sequences with the limit sequence being {1,2,3,...} and since this goes to infinity (as in the sum of the reciprocals diverges), so does the limit of the sequences. However, I am trying to solve the problem for arithmetic progressions of length three, so this problem doesn't arise.
I have found that (using a shanks transformation of the series) that the series converges to about 3.0079
You didn't answer my question.
I gave an example where a set of sequences all converge, but there was no upper bound on what they converged to. How do you know that this can't be the case here?
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
tomtom2357 wrote:if the conjecture is in fact true, then no matter what sequence you have that has no arithmetic progression, it will converge, and it will converge to a number at or below a specific constant. I will call this (hypothetical) constant c.
Why do you think this?tomtom2357 wrote:Now if c doesn't exist, then there is a sequence with no arithmetic progression of length 3 whose reciprocals diverge
Treatid basically wrote:widdout elephants deh be no starting points. deh be no ZFC.
mike-l wrote:Why are you taking limits?
You are talking about the set { {1/a_n} | a_n is an integer, there is no arithmetic progression of length 3 among the a_n}. You are claiming that if it's true that all members of this set have convergent sums, then the sums are bounded above.
I am talking about the set { {1/r^k} | r > 1 }. Now this set happens to have no arithmetic progressions of length 3 as well for all but countably many, so we can ignore those. So aside from the r^k need not be integers, this is similar to your set. Every member of this set converges, but the sums are not bounded. So it's not true in general that if a collection of sequences all converge, then their sums has an upper bound. The existence of such an upper bound is a stronger condition than the convergence.
So you think the set {1,2,4,5,10,...} has a reciprocal sum a little over 3.
Why does this tell you anything about the sequence {1,3,4,6,10,...}?
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
mike-l wrote:Again, a_n converging is a stronger condition than your conjecture. In fact, that simply finds the c you were talking about before if it does.
Both methods you've suggested would prove your sub conjecture, but you haven't offered any idea on how to actually prove either of them.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
tomtom2357 wrote:mike-l wrote:Again, a_n converging is a stronger condition than your conjecture. In fact, that simply finds the c you were talking about before if it does.
Both methods you've suggested would prove your sub conjecture, but you haven't offered any idea on how to actually prove either of them.
All I need to do is prove a lower bound on the series, such as a_{n}>=x^1.5, that would prove its convergence.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.
gmalivuk wrote:Why do you think this?tomtom2357 wrote:Now if c doesn't exist, then there is a sequence with no arithmetic progression of length 3 whose reciprocals diverge
Yes, finding c would prove the conjecture, but the conjecture could also be true even if there is no such c, just as 1/r^n converges for all r<1 but can have an arbitrarily high limit by choosing r arbitrarily close to 1.
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